Nelson Calculus and Vectors Answer Key PDF

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Answers Chapter 1

12.

a. b. 8

Review of Prerequisite Skills, pp. 2–3 1.

2.

3. 4.

5. 6. 7.

8.

9.

10. 11.

a. ⫺3

c. 4

b. ⫺2

d. ⫺4

a. y ⫽ 4x ⫺ 2 b. y ⫽ ⫺2x ⫹ 5 6 c. y ⫽ 1x ⫹ 12 ⫹ 6 5 d. x ⫹ y ⫺ 2 ⫽ 0 e. x ⫽ ⫺3 f. y ⫽ 5 a. ⫺1 c. ⫺9 b. 0 d. 144 5 a. ⫺ c. 0 52 5 3 b. ⫺ d. 13 52 a. 6 c. 9 b. 兹3 d. 兹6 1 a. ⫺ c. 5 e. 106 2 b. ⫺1 d. 1 a. x 2 ⫺ 4x ⫺ 12 b. 15 ⫹ 17x ⫺ 4x 2 c. ⫺x 2 ⫺ 7x d. ⫺x 2 ⫹ x ⫹ 7 e. a 3 ⫹ 6a 2 ⫹ 12a ⫹ 8 f. 729a 3 ⫺ 1215a 2 ⫹ 675a ⫺ 125 a. x1x ⫹ 12 1x ⫺ 12 b. 1x ⫹ 32 1x ⫺ 22 c. 12x ⫺ 32 1x ⫺ 22 d. x1x ⫹ 12 1x ⫹ 12 e. 13x ⫺ 42 19x 2 ⫹ 12x ⫹ 16 2 f. 1x ⫺ 12 12x ⫺ 32 1x ⫹ 22 a. 5x苸R 0 x ⱖ ⫺56 b. 5x苸R6 c. 5x苸R 0 x ⫽ 16 d. 5x苸R 0 x ⫽ 06 1 e. e x苸R 0 x ⫽ ⫺ , 3 f 2 f. 5x苸R 0 x ⫽ ⫺5, ⫺2, 1 6 a. 20.1 m>s b. 10.3 m >s a. ⫺20 L>min b. about ⫺13.33 L>min c. The volume of water in the hot tub is always decreasing during that time period, a negative change.

622 A n s w e r s

7.

4

e. ⫺4.1 1 f. ⫺ 2

x 0

–2

2

4

6

–4

Section 1.2, pp. 18–21

–8

1.

m ⫽ ⫺8 c. ⫺8

2.

Section 1.1, p. 9 1.

2.

3.

4.

d.

b. 10 ⫺ 3兹10

e.

c. 5 ⫺ 2兹6

f.

a.

c.

6.

d.

a. 兹5 ⫹ 兹2

b.

5.

3.

a. 2 兹3 ⫹ 4 d. b. 兹3 ⫺ 兹2 e. c. ⫺2兹3 ⫹ 兹2 f. 兹6 ⫹ 兹10 a. c. 2 b. 兹6 ⫺ 3

12 兹15 ⫹ 15 兹10 2 f. 5 ⫹ 2兹6 1 a. 兹a ⫹ 2 1 b. 兹x ⫹ 4 ⫹ 2 1 c. 兹x ⫹ h ⫹ 兹x e. ⫺

y

3兹3 ⫺ 兹2 兹2 ⫹ 兹5 ⫺ 兹5 ⫺ 2兹2 4 ⫹ 兹6 2 3兹10 ⫺ 2 4 4 ⫺ 2兹5 11 兹6 ⫺ 16 47 35 ⫺ 12 兹6 19

5 1 c. ⫺ 3 3 1 7 a. ⫺ b. ⫺ 3 13 a. 7x ⫺ 17y ⫺ 40 ⫽ 0 y 4 a. 3

b. ⫺

2 x –2

0

2

4

6

–2

–4

1

b.

y ⫽ 8x ⫹ 6

兹5 ⫹ 1 ⫺7

8

y

2 ⫹ 3兹2 1

12 ⫺ 5兹5 a. 8兹10 ⫹ 24 b. 8兹10 ⫹ 24 c. The expressions are equivalent. The radicals in the denominator of part a. have been simplified in part b. a. 兹6 ⫹ 2 9 兹2 ⫹ 2 兹3 b. 25 c. 2兹2 ⫹ 兹6 12 ⫹ 5兹6 d. 2

4 x –4

–2

0

2

4

–4

–8

NEL

7.

c. 3x ⫺ 5y ⫺ 15 ⫽ 0 y 4

P

2 x –2

0

2

4

6

–2

–4

Q (3, 27)

19

(2, 8)

(2.5, 15.625)

15.25

(2, 8)

(2.1, 9.261)

12.61

(2, 8)

(2.01, 8.120 601)

12.0601

(2, 8)

(1, 1)

(2, 8)

(1.5, 3.375)

9.25

(2, 8)

(1.9, 6.859)

11.41

(2, 8)

(1.99, 7.880 599)

11.9401

b. c. d. e. f.

2

2

4

13.

12

6

Take values of x close to the point, y then determine ¢ ¢x .

14.

7

12 12 ⫹ 6h ⫹ h 2 12 They are the same.

x 0

Slope of Line PQ

(2, 8)

d. x ⫽ 5 y 4

–2

Semi-circle y ⫽ 兹25 ⫺ x 2 S centre , rad 5, 10, 02 yⱖ 0 OA is a radius. The slope of OA is 43 . The slope of tangent is ⫺34 .

a.

Since the tangent is horizontal, the slope is 0. 15. 3x ⫺ y ⫺ 8 ⫽ 0 16. 3x ⫹ y ⫺ 8 ⫽ 0 17. a. 13, ⫺22 b. 15, 62 c. y ⫽ 4x ⫺ 14 d. y ⫽ 2x ⫺ 8 e. y ⫽ 6x ⫺ 24

y

8

–2

18.

a. undefined

4 –4 4.

5.

6.

a. 75 ⫹ 15h ⫹ h 2 b. 108 ⫹ 54h ⫹ 12h 2 ⫹ h 3 1 c. ⫺ 1⫹h d. 6 ⫹ 3h ⫺3 e. 414 ⫹ h2 1 f. 4 ⫹ 2h 1 a. 兹16 ⫹ h ⫹ 4 h⫹5 b. 兹h 2 ⫹ 5h ⫹ 4 ⫹ 2 1 c. 兹5 ⫹ h ⫹ 兹5 a. 6 ⫹ 3h b. 3 ⫹ 3h ⫹ h 2 1 c. 兹9 ⫹ h ⫹ 3

x –4

0

–2

2

P

4

–4 8.

5 1 4

c. 12 5 c. 6

a. ⫺12 1 a. 2

b.

10.

a. ⫺2

b. ⫺

11.

a. 1

9.

b.

1 2

d.

c. ⫺

b. 0

1 25

1 6

P

3 e. ⫺ 4 1 f. ⫺ 6

b. ⫺1 c. 9

c. about ⫺2.5

12.

P y 8

4

A x

–4

NEL

0

4

8

Answer s

623

1 The slope of the tangents ata ⫽ 2 is 1 ⫺1 ⫽ MP and at a ⫽ ⫺2 is 1 ⫽ Mq; mp MP ⫽ ⫺1 and mqMq ⫽ ⫺1. Therefore, the tangents are perpendicular at the points of intersection. 24. y ⫽ ⫺11x ⫹ 24 25. a. 8a ⫹ 5 b. 10, ⫺22 c. 1⫺5, 732

d. about 1

P

e. about ⫺

7 8

Section 1.3, pp. 29–31 1. 2.

P

19. 20. 21. 22.

23.

f. no tangent at point P 5 ⫺ 4 500 papers/year (2, 4) 26 28 26 a ⫺2, b , a ⫺1, b , a 1, ⫺ b , 3 3 3 28 a 2, ⫺ b 3 1 y ⫽ x 2 and y ⫽ 2 ⫺ x 2 1 x2 ⫽ ⫺ x2 2 1 2 x ⫽ 4 1 1 x ⫽ or x ⫽ ⫺ 2 2 The points of intersection are

1 1 1 PQ12 , 4 R and Q Q⫺ 2 , 4 R. Tangent to y ⫽ x: 1a ⫹ h2 2 ⫺ a 2 m ⫽ lim h h S0 2ah ⫹ h 2 ⫽ lim h h S0 ⫽ 2a 1 The slope of the tangent at a ⫽ 2 is 1 1 ⫽ mp and at a ⫽ ⫺2 is ⫺1 ⫽ mq.

Tangents to y ⫽ 21 ⫺ x 2:

c 12 ⫺ 1a ⫹ h2 2 d ⫺ c 2 ⫺ a 2 d 1

m ⫽ lim h S0

⫺2ah ⫺ h 2 ⫽ lim h h S0 ⫽ ⫺2a

624 A n s w e r s

h

3.

4.

0 s or 4 s a. Slope of the secant between the points 12, s1222 and 1 9, s 1 922 b. Slope of the tangent at the point 1 6, s 1 622 Slope of the tangent to the function with equation y ⫽ 兹x at the point 14, 22 a. A and B b. greater; the secant line through these two points is steeper than the tangent line at B. c. y y = f (x) B A

5.

12. 13. 14.

15.

16. 17.

18.

1

y ⫺ a1 ⫽ ⫺a 2 1x ⫺ a 2 , or 1 y ⫽ ⫺ 2 x ⫹ a2. The intercepts are 1

Q0, a2R and 1⫺2a, 02 . The tangent line a

and the axes form a right triangle with 2 a

x

Speed is represented only by a number, not a direction. 6. Yes, velocity needs to be described by a number and a direction. Only the speed of the school bus was given, not the direction, so it is not correct to use the word “velocity.” 7. a. first second ⫽ ⫺ 5 m> s, third second ⫽ ⫺ 25 m>s, eighth second ⫽ ⫺ 75 m>s b. ⫺55 m>s c. ⫺20 m> s 8. a. i. 72 km>h ii. 64.8 km> h iii. 64.08 km> h b. 64 km> h c. 64 km> h 9. a. 15 terms b. 16 terms> h 1 10. a. ⫺ mg >h 3 b. Amount of medicine in 1 mL of blood being dissipated throughout the system 1 11. s> m 50

triangle is 12 Q2a R 12a 2 ⫽ 2.

legs of length and 2a. The area of the

19.

E

The coordinates of the point are Qa, a1R. The slope of the tangent is ⫺a 2. The equation of the tangent is

C D

km 5 12 2⫺s; 0 °C m >s a. $4800 > ball b. $80 per c. 0 6 x 6 80 a. 6 b. ⫺1 1 c. 10 $1 162 250 year > a. 75 m b. 30 m s c. 60 m >s d. 14 s >

C1x2 ⫽ F ⫹ V1x2 C1x ⫹ h2 ⫽ F ⫹ V1x ⫹ h2 Rate of change of cost is C1x ⫹ h2 ⫺ C1x2 lim h S0 h V 1x ⫹ h2 ⫺ V 1x2 ⫽ lim , h h S0 which is independent of F ⫺ (fixed costs) 20. 200p m2 >m 21. Cube of dimensions x by x by x has volume V ⫽ x 3. Surface area is 6x 2. V¿ 1x2 ⫽ 3x 2 ⫽ 21 surface area.

22.

a. 80p cm2>unit of time b. ⫺100p cm3>unit of time

Mid-Chapter Review, pp. 32–33 1.

a. 3 b. 37

2.

a.

c. 61 d. 5

6兹3 ⫹ 兹6 3

6 ⫹ 4兹3 3 51 兹7 ⫹ 42 c. ⫺ 9 b.

d. ⫺213 ⫹ 2兹32

NEL

e.

10 兹3 ⫺ 15 2

f. ⫺ 3.

a.

7.

3 兹21 2 兹3 ⫹ 5 2 13 2

8.

3

b. ⫺9

d. 1 6

a.

b.

兹316 ⫹ 兹22 9 c. ⫺ 51 兹7 ⫹ 42 13 d. ⫺ 3 兹21 2 兹3 ⫹ 5 2 1 e. ⫺ 1 兹3 ⫹ 兹72 1 f. 12兹3 ⫺ 兹72 2 a. x ⫹ y ⫺ 6 ⫽ 0 3

c. d. 9. a. b. 10. a. b. 11. a. b. c. d. 12. a. b.

b. x ⫺ y ⫹ 5 ⫽ 0 c. 4x ⫺ y ⫺ 2 ⫽ 0 5.

d. x ⫺ 5y ⫺ 9 ⫽ 0 ⫺2

6.

a.

1⫺1, 1 2 1⫺2, 6 2

Q

1⫺1, 1 2 1⫺1.5, 3.25 2

1⫺1, 1 2 1⫺1.1, 1.41 2

1 ⫺1, 1 2 1 ⫺1.01, 1.0401 2

1. Slope of Line PQ

⫺5

3.

⫺4.5 ⫺4.1

1 ⫺1, 1 2 1 ⫺1.001, 1.004 001 2 ⫺4.001 P

1⫺1, 1 2 10, ⫺2 2

Q

1⫺1, 1 2 1⫺0.5, ⫺0.75 2

1⫺1, 1 2 1⫺0.9, 0.61 2

1 ⫺1, 1 2 1 ⫺0.99, 0.960 1 2

i. 36 km >h ii. 30.6 km>h iii. 30.06 km h > appears to approach velocity of car 30 km h 16h ⫹ >30 2 km>h 30 km >h ⫺4 ⫺12 ⫺2000 L>min ⫺1000 L>min ⫺9x ⫹ y ⫹ 19 ⫽ 0 8x ⫹ y ⫹ 15 ⫽ 0 4x ⫹ y ⫹ 8 ⫽ 0 ⫺2x ⫹ y ⫹ 2 ⫽ 0 ⫺3x ⫹ 4y ⫺ 25 ⫽ 0 3x ⫹ 4y ⫹ 5 ⫽ 0

Section 1.4, pp. 37–39

2. P

⫺4.01

Slope of Line PQ

⫺3 ⫺3.5 ⫺3.9

1 ⫺1, 1 2 1 ⫺0.999, 0.996 0012 ⫺3.999

NEL

1 c. ⫺ 4

5兹2

b.

4.

a. ⫺3

4.

⫺3.99

5. 6.

b. ⫺4 c. h ⫺ 4 d. ⫺4

7.

e. The answers are equal.

8.

72 b. p 99 Evaluate the function for values of the independent variable that get progressively closer to the given value of the independent variable. a. A right-sided limit is the value that a function gets close to as the values of the independent variable decrease and get close to a given value. b. A left-sided limit is the value that a function gets close to as the values of the independent variable increase and get close to a given value. c. A (two-sided) limit is the value that a function gets close to as the values of the independent variable get close to a given value, regardless of whether the values increase or decrease toward the given value. a. ⫺5 d. ⫺8 b. 10 e. 4 c. 100 f. 8 1 a. 0 c. ⫺1 b. 2 d. 2 a. 2 b. 1 c. does not exist a. 8 b. 2 c. 2 a.

9.

5 y

6

4

2 x –4 10.

0

–2

2

a. 0

d. ⫺

b. 0

e.

4 1 2

1 5

c. 5 f. does not exist; substitution causes division by zero, and there is no way to remove the factor from the denominator. 11. a. does not exist c. 2 b. 2 d. does not exist 12. Answers may vary. For example: a. 6

y

4 2 x –8 –6 –4 –2 0 –2

2

4

6

8

2

4

6

8

2

4

6

8

–4

b. 6

y

4 2 x –8 –6 –4 –2

0 –2 –4

c. 6

y

4 2 x –8 –6 –4 –2

0 –2 –4

Answer s

625

d.

6

y

8.

4 2

a.

1 12

d.

b. ⫺27

e.

1 c. 6

f.

a. 0

d.

b. 0

e.

c. ⫺1

f.

x 0 –8 –6 –4 –2 –2

2

4

6

8

–4

9.

13. m ⫽ ⫺3; b ⫽ 1 14. a ⫽ 3, b ⫽ 2, c ⫽ 0 15. a. y 10 8

10.

6

d. exists

1 2 1 12 1 12 1 2 2x 1 32

y 4

2 x –4

0

–2

2

–2

a. does not exist

4

y

2

–4

2 x –4 –2 0 –2

2

4

6

8

10

lim 13 ⫹ x2 and lim 1x ⫹ 32 have the

Section 1.5, pp. 45–47 x S2

x S2

same value, but lim 3 ⫹ x does not.

3.

4.

5. 6.

7.

Since there are no brackets around the expression, the limit only applies to 3, and there is no value for the last term, x. Factor the numerator and denominator. Cancel any common factors. Substitute the given value of x. Yes, if the two one-sided limits have the same value, then the value of the limit is equal to the value of the one-sided limits. If the one-sided limits do not have the same value, then the limit does not exist. a. 1 d. 5p3 b. 1 e. 2 100 c. f. 兹3 9 a. 2 b. 兹2 Since substituting t ⫽ 1 does not make the denominator 0, direct substitution works. 1 ⫺ 1 ⫺ 5 ⫽ ⫺5 5 6⫺1 ⫽ ⫺1 1 a. 4 d. ⫺ 4 1 b. 1 e. 4 1 c. 27 f. ⫺ 兹7

–8

–4

0

4

8

V

⫺40

19.1482

⫺20

20.7908

0

22.4334

20

24.0760

40

25.7186

60

27.3612

80

29.0038

20

–1

20 20

–2

20

b. does not exist

20

y

20

4

2 x –3 –2 –1 0

1

2

3

–2

–4

⌬V 1.6426 1.6426 1.6426 1.6426 1.6426 1.6426

¢V is constant; therefore, T and V form a linear relationship. b. V ⫽ 0.082 13T ⫹ 22.4334 V ⫺ 22.4334 c. T ⫽ 0.082 13 0 ⫺ 22.4334 d. lim T ⫽ 0.082 13 v S0⫹ ⫽ ⫺273.145 e. y 50 40

c. does not exist y

30

4

20 10

2

x x

–3 –2 –1 0 –2

1

2

3

–300 –200 –100

0

100 200

x2 ⫺ 4 x S5 f 1x2

12. lim

⫽ –4

626 A n s w e r s

T

x

x S2

2.

a.

⌬T

b. 6; 4 c. 2000 d. about 8.49 years

1.

11.

1

12

lim1x 2 ⫺ 42 lim f 1x2

x S5

x S5

21 ⫽ 3 ⫽7

NEL

13. 14.

a. 27 a. 0

15. 16. 17.

a. 0 ⫺2 does not exist

4.

c. 1

b. ⫺1 b. 0 1 b. 2

y

a. x ⫽ 3 b. x ⫽ 0 c. x ⫽ 0 d. x ⫽ 3 and x ⫽ ⫺3 e. x ⫽ ⫺3 and x ⫽ 2 f. x ⫽ 3 a. continuous for all real numbers b. continuous for all real numbers c. continuous for all real numbers, except 0 and 5 d. continuous for all real numbers greater than or equal to⫺2 e. continuous for all real numbers f. continuous for all real numbers g1x2 is a linear function (a polynomial), and so is continuous everywhere, including x ⫽ 2.

5.

4

2 x –4

0

–2

2

6.

–2

10. no 11. Discontinuous at x ⫽ 2 12. k ⫽ 16 13. a. y 4

2 x –4

0

–2

2

4

–2

–4

7.

1.

2.

3.

b.

y

Section 1.6, pp. 51–53

6

Anywhere that you can see breaks or jumps is a place where the function is not continuous. On a given domain, you can trace the graph of the function without lifting your pencil. point discontinuity

4

c.

2

14.

x –6 –4 –2 0 –2

2

4

6

a. b. c.

y

x S3

8

Yes, the function is continuous everywhere.

4 2

x –2 0 –2

2

15.

(1) A ⫽ B ⫺ 3 (2) 4B ⫺ A ⫽ 6 (if B 7 1, then A 7 ⫺2; if B 6 1, then A 6 ⫺2) 16. a ⫽ ⫺1, b ⫽ 6

8.

hole

4

4

6

y

17.

a. lim ⫺ g1x2 ⫽ ⫺1 x S1

2

jump discontinuity

x S3

Hence, f is not continuous at x ⫽ 3 and also not continuous on ⫺3 6 x 6 8 .

–6

6

Thus, lim f 1x2 ⫽ 4. But, f 132 ⫽ 2.

x S3

–4

10

i. ⫺1 ii. 1 iii. does not exist f is not continuous since lim f 1x2 x S0 does not exist. 2 4 lim⫺ f 1x2 ⫽ 4 ⫽ lim⫹ f 1x2

lim ⫹ g1x2 ⫽ 1 x S1

y

lim g1x2 does not exist.

x

10 8

–4

–2

0

2

b.

4

x S1

4

6 4

g1x2 ¶ xlim S1

y

–2

2

2

x –2 0 –2

2

4

6

–4

x

The function is discontinuous at x ⫽ 0.

infinite discontinuity

9.

y

–4

–2

0

2

4

y

10

–2

4

8 6 4

–4

2

2

g1x2 is discontinuous at x ⫽ 1.

x –1 0 –2 –4

NEL

1

2

3

x

4

vertical asymptote

0

200

400

600

Discontinuities at 0, 100, 200, and 500

Answer s

627

Review Exercise, pp. 56–59 1. 2.

3. 4.

5.

6.

7.

8.

9.

a. ⫺3 b. 7 c. 2x ⫺ y ⫺ 5 ⫽ 0 1 ⫺1 a. c. ⫺ 3 27 1 5 b. d. ⫺ 2 4 a. 2 b. 2 a. m s, m s b. m s c. 1st second m s ⫽ ⫺5 > a. 2nd 0.0601 g ⫽ ⫺15 > second b. ⫺40 6.01 g >min c. ⫺60 6 g min> a. 700 000 t 4 b. 18 ⫻ 10 > t per year c. 15 >⫻ 104 t per year d. 7.5 years a. 10 b. 7; 0 c. t ⫽ 3 and t ⫽ 4 a. Answers may vary. For example: 2

a. 4

y

a. 10

17.

a.

2

b. 18.

x –4

0

–2

2

4

a. b.

–2

–4

10. 11.

c.

b. x ⫽ ⫺1 and x ⫽ 1 c. They do not exist. not continuous at x ⫽ ⫺4 a. x ⫽ 1 and x ⫽ ⫺2 2 b. lim f 1x2 ⫽ , x S1 3 lim f 1x2 does not exist.

1 4

c. ⫺

x S1⫺

x S1 ⫹

e. x S 0⫺ 0x 0 ⫽ ⫺x

0x 0 ⫽ ⫺1 x 0x 0 lim ⫽1 x S0 ⫹ x 0x 0 0x 0 lim ⫽ lim⫺ x S0 ⫹ x x S0 x lim

a. lim f 1x2 does not exist.

x S0⫺

x S0

b. lim g1x2 ⫽ 0 x S0

y

b.

d. The function has a vertical asymptote at x ⫽ 2.

x S⫺2

12.

1 16 1 1 4 c. e. ⫺ 8 兹5 1 1 10a d. f. ⫺ 3 4 The function is not defined for x 6 3, so there is no left-side limit. Even after dividing out common factors from numerator and denominator, there is a factor of x ⫺ 2 in the denominator; the graph has a vertical asymptote at x ⫽ 2. lim f 1x2 ⫽ ⫺5 ⫽ lim f 1x2 ⫽ 2

16.

37 7 lim h1x2 does not exist.

c. lim h1x2 ⫽ x S⫺3

x S⫺3

1

13.

x

1.9

1.99

1.999

2.001